# Question about distance between point and plane

Thanks for reading. My question is about a specific step in the proof of the theorem used to calculate the distance between a point and a plane. The proof proceeds up to this point:

And from here the textbook reads as follows:

Since the point $Q(x_1,y_1,z_1)$ lies in the given plane, its coordinates satisfy the equation of that plane; thus

$$ax_1 + by_1 + cz_1 + d = 0$$

or

$$d = -ax_1 - by_1 - cz_1$$

Substitute this expression in the above equation

Why is this substitution a valid step? I don't understand what "$d$" signifies. And do I simply replace everything above the divide?

Thanks

• What is the equation of the plane? Is not $ax+by+cz+d=0?$ Thus, any point $(x_0,y_0,z_0)$ of the plane satisfies $ax_0+by_0+cz_0+d=0.$ Equivalently, $d=-ax_0-by_0-cz_0.$ – mfl Aug 23 '17 at 20:15
• Thanks for taking the time comment, mfl. Sadly, I still don't understand. Let met try to trace what I do know and what I don't. I do know that ax +by + cz + d = 0 represents an equation for a plane. I do know that (a,b,c) of that equation represents the normal of the plane and that (x,y,z) represents the coordinates of a point on that plane. What I don't know is what the d represents. Also, even though I know that ax1 + by1 +cz1 + d = 0 can be written as d = -ax1 - by1 - cz1, I still don't know why it's relevant here. Why can d = -ax1 - by1 - cz1 replace |a(x0 -x1) + b(y0 - y1) + c(z0 - z1)|? – Tightrope Aug 23 '17 at 21:01
• $d$ doesn't represent $|a(x_0 -x_1) + b(y_0 - y_1) + c(z_0 - z_1)|.$ We have that $|ax_0 +by_0 + cz_0 -ax_1 +by_1 + cz_1|.$ Now $ax_0 +by_0 + cz_0$ depends on the point $(x_0,y_0,z_0)$ but $ax_1 +by_1 + cz_1$ doesn't depend on the chosen point $Q.$ It is the constant $-d.$ Thus you get the desired formula. – mfl Aug 23 '17 at 21:07
• By the way, what's the geometrical meaning of $d?$ If you choose the normal vector $(a,b,c)$ to be a unit vector then $d$ is just the distance from the origin to the plane. – mfl Aug 23 '17 at 21:09
• Thank you mfl, I finally got my head around it. Took me a while. – Tightrope Aug 24 '17 at 12:53

Consider the plane:

$Ax + By + Cz + D = 0$;

Normalized normal vector to this plane:

$\vec n$: $\dfrac{1}{\sqrt{A^2 +B^2+C^2}} (A,B,C).$

Point $P$ in the plane: $(x_0,y_0,z_0)$.

Vector pointing from $P (x_0,y_0,z_0)$ to $Q(x_1,y_1,z_1)$:

$\vec r = (x_1 - x_0, y_1 - y_0, z_1- z_0)$.

Using the scalar product the distance of $Q$ from the plane:

$|\vec n \cdot \vec r| =$

$\dfrac{1}{\sqrt{A^2 + B^2 + C^2}}$

$|(A(x_1-x_0) + B(y_1 - y_0) +$

$C(z_1 - z_0)|$ .

Since $Ax_0 + By_0 + Cz_0 = - D$

substitute in the above equation.

Helps?

As you can see from the equation of a plane, $-d$ is a measure of the distance of the plane from the origin, with sign, but not normalized.
Taking $a(x_0-x_1)+b(y_0-y_1)+c(z_0-z_1)$ is to take the difference of the not normalized distances from the origin, then normalize ( dividing by the modulus of the vector) and take the absolute value.